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[leetcode]Pascal's triangle 2

Given a non-negative index k where k ≤ 33, return the kth index row of the Pascal's triangle.

Note that the row index starts from 0.

分析:

楊輝三角要求第k行(k從0開始)的值,楊輝三角每一行的首尾的值都為1,每一行值的個數依次加一。且第k行(k>=2)的第j個值(1<=j<k)為第k-1行的第j和第j-1個值相加所得。

class Solution {
public:
    vector<int> getRow(int rowIndex) {
        vector<int> res;
        for (int k = 0; k <= rowIndex; k++) {
            vector<int> curr(k+1, 1);//定義一個長度為k+1,值都為1的一維陣列
            for(int j = 1; j < k; j++) {
                curr[j] = res[j-1]+res[j];
            }
            res = curr;
        }

        return res;        
    }
};